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Question

In the given figure, $$AB = AC$$. Then, can we say that  AD bisects angle A
194682_6ef9423c50164dbdbc56cf94423292c3.png


Solution

Given, $$AB = AC$$
$$\angle ABC = \angle ACB$$ (Isosceles triangle property)
Now, in $$\triangle DPB$$ and $$\triangle DQC$$,
$$BD = DC$$ (Given)
$$\angle DBP = \angle DCQ$$ (Proved above)
$$\angle DPB = \angle DQC$$ (Each $$90^{\circ}$$)
thus, $$\triangle DPB \cong \triangle DQC$$ (ASA rule)
Hence, $$PB = QC$$ (By cpct)
We know, $$AB = AC$$
Hence, $$AB - PB = AC - QC$$
$$AP = AQ$$ (I)
Now, In $$\triangle APD$$ and $$\triangle AQD$$,
$$\angle APD = \angle AQD$$ (each $$90^{\circ}$$)
$$ AP = AQ $$ (From I)
$$AD = AD $$ (Common)
Thus, $$\triangle APD \cong \triangle AQD$$ (SAS postulate)
Hence, $$\angle DAP = \angle DAQ$$ (By cpct)
$$AD$$ bisects $$\angle A$$

Mathematics

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