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Question
In quadrilateral ABCD, AB = BC and AD = CD. Show that BD bisects ∠ABC and ∠ADC both.
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Solution
Given: AB = BC and AD = CD in quadrilateral ABCD.
To Prove: BD bisects ∠ABC and BD bisects ∠ADC (i.e., ∠ABD = ∠DBC and ∠ADB = ∠BDC).
Proof (Step-wise):
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Consider triangles △ABD and △CBD.
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AB = BC. ...(Given)
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AD = CD. ...(Given)
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BD = BD. ...(Common side)
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From (2), (3) and (4), the three sides of △ABD are respectively equal to the three sides of △CBD. Hence △ABD ≅ △CBD by SSS congruence.
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By corresponding parts of congruent triangles, ∠ABD = ∠DBC and ∠ADB = ∠BDC. This shows BD bisects ∠ABC and BD bisects ∠ADC. (Compare this use of triangle congruence and CPCTC with the congruence/angle-equality reasoning in the provided geometry notes.)
Therefore, BD bisects both ∠ABC and ∠ADC.
