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in Congruence and Inequalities of Triangles by (31.2k points)
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In figure, ∠A = 35°, ∠ABC = 100° and BD ⊥ AC, prove that ΔBDC is an isosceles triangle.

1 Answer

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Best answer

In ΔABC,

∠A + ∠B + ∠C = 180° (the sum of the interior angles of a Δ is equal to 180°)

35° + 100° + ∠C= 180°

⇒ ∠C = 180° – 135°

⇒ ∠C = 45°

But in ΔABD, BD ⊥ AC

i.e., ∠ADB = 90°

and ∠A = 35° (given)

∠ABD = 180° – 125° = 55°

∠DBC = 100° – 55° = 45°

Now, ∠DCB = ∠DBC = 45° (Proved above)

Hence BDC is an isosceles triangle.

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