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Isosceles Triangle — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSIsosceles Triangle4 Marks
Question
$\triangle ABC$ is an isosceles triangle with $AB = AC$. Another $\triangle BDC$ is drawn with base $BC = BD$ in such a way that $BC$ bisects $\angle B$. If the measure of $\angle BDC$ is $70^\circ $, find the measures of $\angle DBC$ and $\angle BAC.$
✓

Answer

In $\triangle BDC,$
$\angle BDC = 70^\circ $
$BD = BC$
Therefore, $\angle BDC = \angle BCD$
$\Rightarrow \angle BCD = 70^\circ $
Now $\angle BCD + \angle BDC + \angle DBC = 180^\circ $
$70^\circ + 70^\circ + \angle DBC = 180^\circ $
$\angle DBC = 40^\circ $
$\angle DBC =\angle ABC ...(BC$ is the angle bisector$)$
$\Rightarrow \angle ABC = 40^\circ $
In $\triangle ABC,$
Since $AB = AC, \angle ABC = \angle ACB \Rightarrow \angle ACB = 40^\circ $
$\angle ACB + \angle ABC + \angle BAC = 180^\circ $
$40^\circ + 40^\circ + \angle BAC = 180^\circ $
$\angle BAC = 100^\circ$
Hence, $\angle BAC = 100^\circ $ and $\angle DBC = 40^\circ .$

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