Solution

$\mathrm{△}ABC\mathrm{△}ABD$. What can you say about $BC$ and $BD$ ?

In $\mathrm{△}ABC$ and $\mathrm{△}ABD$,

$AC=AD$ (Given)

$\mathrm{\angle }CAB=\mathrm{\angle }DAB(AB$ bisects $\mathrm{\angle }A)$

$AB=AB$ (Common)

$\therefore \mathrm{△}ABC\cong \mathrm{△}ABD$ (By SAS congruence rule)

$\therefore BC=BD(ByCPCT)$

Therefore, BC and BD are of equal lengths.

Solution

In $\mathrm{△}ABD$ and $\mathrm{△}BAC$,

$AD=BC($ Given)

$\mathrm{\angle }\mathrm{\angle }$

$DAB=CBA($ Given)

$AB=BA$ (Common)

$\therefore \mathrm{△}ABD\cong \mathrm{△}BAC$ (By SAS congruence rule)

$\therefore BD=AC$ (By CPCT) And, $\mathrm{\angle }ABD$

$=\widehat{BAC}(ByCPCT)$

Solution

In $\mathrm{△}BOC$ and $\mathrm{△}AOD$,

$\begin{array}{rl}& \mathrm{\angle }\mathrm{\angle }BOC=AOD\text{ (Vertically opposite angles) }\\ & \mathrm{\angle }CBO=DAO(\text{ Each }{90}^{\circ })\\ & BC=AD\text{ (Given) }\\ & \therefore \mathrm{△}BOC\cong \mathrm{△}AOD\text{ (AAS congruence rule) }\\ & \therefore BO=AO(ByCPCT)\\ & \Rightarrow CD\text{ bisects }AB.\end{array}$

Solution

In $\mathrm{△}ABC$ and $\mathrm{△}CDA$,

$\begin{array}{rl}& \mathrm{\angle }BAC=\mathrm{\angle }DCA\text{ (Alternate interior angles, as }p|q)\\ & AC=CA\text{ (Common) }\\ & \mathrm{\angle }\mathrm{\angle }BCA=DAC\text{ (Alternate interior angles, as }\mid |m\text{ ) }\\ & \therefore \therefore \mathrm{△}ABC\mathrm{△}CDA(By\text{ ASA congruence rule) }\\ & \text{ Question 5: }\end{array}$

Solution

In $\mathrm{△}APB$ and $\mathrm{△}AQB$,

$\begin{array}{rl}& \therefore \therefore APB=AQB(\text{ Each 90) }\\ & \therefore \therefore PAB=QAB(I\text{ is the angle bisector of }\therefore A)\end{array}$

$AB=AB$ (Common)

$\therefore \mathrm{△}APB\therefore \mathrm{△}AQB$ (By AAS congruence rule) $\therefore BP=$

BQ (By CPCT)

rms of $\therefore A$. Or, it can be said that $B$ is equidistant from the a

Solution

It is given that $\therefore BAD=\therefore EAC$

$\therefore BAD+\therefore DAC=\therefore EAC+\therefore DAC$

$\therefore BAC=\therefore DAE$

In $\mathrm{△}BAC$ and $\mathrm{△}DAE,AB=AD$

(Given) $\therefore BAC=$

$\therefore$ DAE (Proved above)

$AC=AE($ Given)

$\therefore \mathrm{△}BAC\therefore \mathrm{△}DAE$ (By SAS congruence rule)

$\therefore BC=DE(ByCPCT)$

Solution

It is given that EPA $=$ DPB

$\therefore \therefore EPA+D^{\prime }PE=DPB+DPE$

$\therefore \therefore$ DPA $=$ ÉPB

In $\sqrt{\mathrm{\Delta }}\sqrt{\mathrm{\Delta }}$ EBP,

$\therefore$ DAP $=$ 岸 (Given)

$AP=BP(P$ is mid-point of $AB)$

$\therefore \therefore$ DPA $=$ EPB (From above)

$\therefore \therefore \mathrm{△}$ DAP $\mathrm{△}$ EBP (ASA congruence rule)

$\therefore AD=BE(ByCPCT)$

Solution

(i) In $\mathrm{△}AMC$ and $\mathrm{△}BMD$,

$AM=BM$ ( $M$ is the mid-point of $AB$ )

$\therefore$ AMC $=\therefore$ BMD (Vertically opposite angles)

$CM=DM$ (Given)

$\therefore \mathrm{△}AMC\therefore \mathrm{△}BMD$ (By SAS congruence rule)

$\therefore AC=BD(ByCPCT)$ And,

$\therefore$ ACM $=:BDM(By$ CPCT) ii)

$\therefore ACM=‘BDM($

However, AेंCM and ¿BDM are alternate interior angles.

Since alternate angles are equal,

It can be said that $DB|AC$

$\begin{array}{rl}& \therefore DBC+\therefore ACB={180}^{\circ }\text{ (Co-interior angles) }\\ & \therefore \therefore DBC+{90}^{\circ }={180}^{\circ }\\ & \therefore \therefore DBC={90}^{\circ }\end{array}$

(iii) In $\mathrm{△}DBC$ and $\mathrm{△}ACB$,

$DB=AC$ (Already proved)

$\therefore DBC=\therefore ACB(.$ Each 90 $.{}^{\circ })$

$BC=CB$ (Common)

$\therefore \mathrm{△}DBC\mathrm{△}ACB$ (SAS congruence rule) iv)

$\mathrm{△}DBC\mathrm{△}ACB($

$\therefore AB=DC(ByCPCT)$

$\therefore AB=2CM$

$\therefore CM=\frac{1}{2}AB$