In the adjoining figure, AB = 10 cm, BC =15 cm AD : DC = 2 : 3, then find $∠$ ABC.

[3 Marks]

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Solution

Clearly, $\frac{A D}{D C}$ = $\frac{2}{3}$ and $\frac{A B}{B C}$ = $\frac{10}{15}$ = $\frac{2}{3}$

So, $\frac{A D}{D C}$ = $\frac{A B}{B C}$ [1 Mark]

Thus, BD divides AC in the ratio of the other two sides.

By Converse of Angular Bisector theorem,
BD is the bisector of $∠$ B.
$∴$ $∠$ ABC = 2 ( $∠$ CBD) [1 Mark]

Now, $∠$ CBD = $180^{\circ}$ - ( $130^{\circ}$ + $30^{\circ}$ ) ....(Angle Sum Property)
= $20^{\circ}$ [0.5 Marks]

But , $∠$ ABC = 2 ( $∠$ CBD)
= 2 $\times$ 20
= $40^{\circ}$ [0.5 Marks]

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In the adjoining figure, AB = 10 cm, BC =15 cm AD : DC = 2 : 3, then find ∠ABC. [3 Marks]

In the adjoining figure, AB = 10 cm, BC =15 cm AD : DC = 2 : 3, then find $∠$ ABC.

[3 Marks]