Question

(a) In the given figure, prove that:


(i) $\Delta ABC\cong \Delta DCB$

(ii) AC = DB

(b) In the figure, it is given that $\angle$A = ${90}^{\circ }$, AB = AC, and D is the mid point of BC. Find $\angle$ADC.

[4 MARKS]

Answer 1

Each Part: 2 Marks

(a)


(i) In $\Delta ABC$ and $\Delta DCB$

$AB=DC$ [Given]

$\angle ABC=\angle DCB={90}^{\circ }$ [Given]

$BC=BC$ [Common side]

$\Rightarrow \Delta ABC\cong \Delta DCB$ [SAS congruency criteria]

(ii) $\therefore AC=DB$ [Corresponding parts of congruent triangles]


(b)

AB = AC (Given)
BD = DC (D is mid point of BC)
AD = AD (Common side)

Therefore, $△$ADB $\cong$ $△$ADC [SSS congruency criteria]

thus, $\angle$ADB = $\angle$ADC (c.p.c.t.) ...(1)

but, $\angle$ADB + $\angle$ADC = ${180}^{\circ }$ [linear pair]

$\therefore \angle$ADC + $\angle$ADC = ${180}^{\circ }$

$\Rightarrow 2\angle$ADC = ${180}^{\circ }$

$\Rightarrow \angle ADC=\frac{180°}{2}$

$\Rightarrow \angle ADC=90°$