Question

8. In the following figure;
AC = CD; angle BAD = 110° and angle ACB = 74º.
Prove that : BC > CD.​

Answer 1

Answer with Step-by-step explanation:

Angle BAD=110 degrees

In triangle ACD

AC=CD

Angle ADC=Angle CAD

Angle made by two equal sides are equal

Angle ACD+angle ACB=180 degrees

Reason: Linear pair sum

Angle ACD=180-angle ACB

$\angle ACD=180-74=106$ degrees

$\angle ACD+\angle ADC+\angle CAD=180$degrees

Reason: Triangle angles sum property

$\angle CAD+\angle CAD+106=180$

$2\angle CAD=180-106=74$

$\angle CAD=\frac{74}{2}=37^{\circ}$

In triangle ABC

Angle BAC=110-angle CAD=110-37=73degrees

Angle BAC>Angle CAD

BC>CD

When angle a > angle b

Side opposite to a is greater than the side opposite to b

Hence, proved.

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