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Triangles — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSTriangles3 Marks
Question
In the given figure, ABCD is a square and $\triangle PAB$ is an equilateral triangle.
(i) Prove that $\triangle APD \cong \triangle BPC$.
(ii) Show that $\angle DPC =15^{\circ}$.
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Answer

[Hint.
(i) $\angle P A D=90^{\circ}+60^{\circ}=150^{\circ}$ and $\angle P B C=90^{\circ}+60^{\circ}=150^{\circ}$.
(ii) $\angle P A D=150^{\circ}$ and $A P=A D \Rightarrow \angle A P D=\angle A D P=15^{\circ}$.]

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