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Q1

Given. In quadrilateral $ACBD$, $AC=AD$ and $AB$ bisects $\mathrm{\angle }A$.
To prove. $\mathrm{△}ABC\cong \mathrm{△}ABD$. What about $BC$ and $BD$?

Proof. Since $AB$ bisects $\mathrm{\angle }A$, $\mathrm{\angle }CAB=\mathrm{\angle }BAD$. Also $AC=AD$ and $AB=AB$(common). Thus in $\mathrm{△}ABC$ and $\mathrm{△}ABD$ we have two sides and the included angle equal → SAS, so $\mathrm{△}ABC\cong \mathrm{△}ABD$.
By CPCT, corresponding parts are equal so $BC=BD$.

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Q2

Given. Quadrilateral $ABCD$ with $AD=BC$ and $\mathrm{\angle }DAB=\mathrm{\angle }CBA$
To prove. (i) $\mathrm{△}ABD\cong \mathrm{△}BAC$ (ii) $BD=AC$ (iii) $\mathrm{\angle }ABD=\mathrm{\angle }BAC$.

Proof. Consider $\mathrm{△}ABD$ and $\mathrm{△}BAC$. We have $AD=BC$ (given), $AB=BA$ (common) and the included angles $\mathrm{\angle }DAB=\mathrm{\angle }CBA$ (given). So by SAS the triangles are congruent. By CPCT, $BD=AC$ and $\mathrm{\angle }ABD=\mathrm{\angle }BAC$.

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Q3

Given. $AD$ and $BC$ are equal perpendiculars to the line segment $AB$. (Fig. 7.18)
To prove. $CD$ bisects $AB$.

Proof. Let $O$ be the intersection of $CD$ and $AB$. In $\mathrm{△}AOD$ and $\mathrm{△}BOC$:
$\mathrm{\angle }OAD=\mathrm{\angle }OBC={90}^{\circ }$, $AD=BC$ (given), and $\mathrm{\angle }AOD=\mathrm{\angle }BOC$ (vertically opposite). Hence AAS → $\mathrm{△}AOD\cong \mathrm{△}BOC$. By CPCT, $AO=O\mathrm{B }$Thus $CD$ bisects $AB$(and $O$ is midpoint).

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Q4

Given. Lines $l$ and $m$ are parallel and are intersected by another pair of parallel lines $p$ and $q$ (see Fig. 7.19).
To prove. $\mathrm{△}ABC\cong \mathrm{△}CDA$

Proof. From parallel-lines properties we get corresponding/alternate angles: $\mathrm{\angle }BAC=\mathrm{\angle }DCA$ and $\mathrm{\angle }BCA=\mathrm{\angle }DAC$. Also $AC=CA$ (common). Thus by ASA, $\mathrm{△}ABC\cong \mathrm{△}CDA$.

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Q5

Given. Line $l$ bisects $\mathrm{\angle }A$. $B$ is any point on $l$. From $B$ drop perpendiculars $BP$ and $BQ$ to the two arms of $\mathrm{\angle }A$ (Fig. 7.20).
To prove. (i) $\mathrm{△}APB\cong \mathrm{△}AQB$  (ii) $BP=BQ$

Proof. In $\mathrm{△}APB$ and $\mathrm{△}AQB$: $\mathrm{\angle }APB=\mathrm{\angle }AQB={90}^{\circ }$, $\mathrm{\angle }PAB=\mathrm{\angle }BAQ$(because $l$ bisects $\mathrm{\angle }A$), and $AB=AB$ (common). So by AAS, triangles are congruent. By CPCT, $BP=BQ$; i.e., $B$ is equidistant from the arms of $\mathrm{\angle }A$.

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Q6

Given. In Fig. 7.21, $AC=AE$, $AB=AD$ and $\mathrm{\angle }BAD=\mathrm{\angle }EAC$
To prove. $BC=DE$

Proof. From $\mathrm{\angle }BAD=\mathrm{\angle }EAC$, add $\mathrm{\angle }DAC$ to both sides to get $\mathrm{\angle }BAC=\mathrm{\angle }DAE$. In $\mathrm{△}BAC$ and $\mathrm{△}DAE$: $AB=AD$, $AC=AE$ (given) and $\mathrm{\angle }BAC=\mathrm{\angle }DAE$. Thus SAS gives $\mathrm{△}BAC\cong \mathrm{△}DAE$. By CPCT, $BC=DE$

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Q7

Given. $AB$ is a line segment, $P$ its midpoint. Points $D$ and $E$ lie on the same side of $AB$ with $\mathrm{\angle }BAD=\mathrm{\angle }ABE$ and $\mathrm{\angle }EPA=\mathrm{\angle }DPB$ (Fig. 7.22).
To prove. (i) $\mathrm{△}DAP\cong \mathrm{△}EBP$. (ii) $AD=BE$

Proof. Given $\mathrm{\angle }EPA=\mathrm{\angle }DPB$. Add $\mathrm{\angle }DPE$ to both sides to get $\mathrm{\angle }DPA=\mathrm{\angle }EPB$. Also $\mathrm{\angle }DAP=\mathrm{\angle }EBP$ (given) and $AP=BP$ (since $P$ is midpoint). So in $\mathrm{△}DAP$ and $\mathrm{△}EBP$ we have two angles equal and included side equal → ASA. Thus the triangles are congruent and by CPCT, $AD=BE$.

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Q8

Given. $\mathrm{△}ABC$ right-angled at $C$. $M$ is midpoint of hypotenuse $AB$. Join $CM$ and extend it to $D$ so that $DM=CM$. Join $BD$. (Fig. 7.23)
To prove.

1. $\mathrm{△}AMC\cong \mathrm{△}BMD$

2. $\mathrm{\angle }DBC$ is a right angle.

3. $\mathrm{△}DBC\cong \mathrm{△}ACB$

4. $CM=\frac{1}{2}AB$

Proof.
(i) In $\mathrm{△}AMC$ and $\mathrm{△}BMD$: $AM=BM$ (M is midpoint of $AB$), $CM=DM$ (given), and $\mathrm{\angle }AMC=\mathrm{\angle }BMD$ (vertically opposite). Hence SAS gives $\mathrm{△}AMC\cong \mathrm{△}BMD$.

(ii) From (i) we get $\mathrm{\angle }ACM=\mathrm{\angle }BDM$. But those are alternate interior angles for lines $AC$ and $BD$cut by transversal through $M$, so $AC\parallel BD$. Since $\mathrm{\angle }ACB={90}^{\circ }$, with $AC\parallel BD$ the angle $\mathrm{\angle }DBC$ (formed by $DB$ and $BC$) equals ${90}^{\circ }$. Thus $\mathrm{\angle }DBC$ is a right angle.

(iii) Now $BD\parallel AC$ and $BC$ is common, so in $\mathrm{△}DBC$ and $\mathrm{△}ACB$: $DB=AC$(corresponding), $\mathrm{\angle }DBC=\mathrm{\angle }ACB$ (both ${90}^{\circ }$), and $BC=CB$. So SAS (or RHS) gives $\mathrm{△}DBC\cong \mathrm{△}ACB$.

(iv) From standard result (midpoint of hypotenuse) or from congruence in (i): $AM=BM=CM$. Since $AM+MB=AB$, $AB=2\cdot AM$ and $AM=CM$ so $CM=\frac{1}{2}AB$