Same chapter4 marksDifficulty: Hard
In triangle ABC, right angled at B, BD is the perpendicular from B to AC. Prove that triangle ADB is similar to triangle ABC, and hence show that AB2 = AD × AC.
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4 marksDifficulty: Hard
In triangle ABC the bisector of angle A meets BC at D. Prove that BD/DC = AB/AC.
Original Khojo Papers practice question — not from a past board paper.
Answer
Drawing a line through C parallel to AD to meet BA produced at E gives triangle BAD similar to triangle BEC, and AE = AC, from which BD/DC = AB/AC.
No citable source has been recorded for this record. Treat it as practice material, not as fact.
Explanation
Since AD is parallel to CE, angle ACE = angle DAC and angle AEC = angle BAD. As AD bisects angle A these two angles are equal, so triangle ACE is isosceles with AE = AC. Applying the basic proportionality theorem in triangle BEC gives BD/DC = BA/AE = AB/AC.
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