Question

In triangle ABC, D is a point in AB such that AC = CD = DB. If $\angle B={28}^{\circ }$, then the measure of angle ACD is

• A. ${54}^{\circ }$
• B. ${64}^{\circ }$
• C. ${72}^{\circ }$
• D. ${84}^{\circ }$

Answer 1

The correct option is B

${64}^{\circ }$

We see that triangle DBC is an isoceles triangle.

As, Side CD = Side DB, we must have, $\angle DBC=\angle DCB$ (In a triangle, if two sides are equal, then the angles opposite to these sides are also equal).

And $\angle B=\angle DBC=\angle DCB={28}^{\circ }$

As the sum of all angles of a triangle is ${180}^{\circ }$,

$\angle DBC+\angle DCB+\angle BDC={180}^{\circ }$

$⟹{28}^{\circ }+{28}^{\circ }+\angle BDC={180}^{\circ }$

Therefore $\angle BDC={124}^{\circ }$

We know that the sum of two non-adjacent interior angles of a triangle is equal to the exterior angle.

$⟹\angle DBC+\angle DCB=\angle ADC$

i.e. ${28}^{\circ }+{28}^{\circ }=\angle ADC$

$⟹\angle ADC={56}^{\circ }$

Now, $△$ACD is an isosceles triangle with AC =DC.

$⟹\angle ADC=\angle DAC={56}^{\circ }$

Since sum of all angles of a triangle is ${180}^{\circ }$,

$\angle ADC+\angle DAC+\angle DCA={180}^{\circ }$

i.e. ${56}^{\circ }+{56}^{\circ }+\angle DCA={180}^{\circ }$

$⟹\angle DCA={64}^{\circ }=\angle ACD$