Question 2
In figure, ∠ ACB = 90° and CD ⊥ AB. Prove that BC2/AC2=BD/AD
Given:- ΔCAB,
∠ ACB = 90°
And CD ⊥ AB
To Prove :- BC2/AC2=BD/AD
Proof :
From theorem 6.7,
If a perpendicular is drawn from the vertex of the right angle to
the hypotenuse then triangles on both sides of the
Perpendicular are similar to the whole triangle and to each other
So, ∆ 𝐴𝐶𝐷 ~ ∆ 𝐴𝐵𝐶
& ∆ 𝐵𝐶𝐷 ~ ∆ 𝐵𝐴𝐶.
From (1)
∆ 𝐴𝐶𝐷 ~ ∆ 𝐴𝐵𝐶
If two triangles are similar ,
then the ratio of their corresponding sides are equal
𝐴𝐶/𝐴𝐵=𝐴𝐷/𝐴𝐶
AC ×𝐴𝐶=𝐴𝐵×𝐴𝐷
AC2 = AB ×𝐴𝐷
Similarly, from (2)
∆ 𝐵𝐶𝐷 ~ ∆ 𝐵𝐴𝐶
If two triangles are similar ,
then the ratio of their corresponding sides are equal
𝐵𝐶/𝐵𝐴=𝐵𝐷/𝐵𝐶
BC × BC = BA × BD
BC2 = BA × BD
Divide equation ((4))/((3))
𝐵𝐶2/𝐴𝐶2=(𝐵𝐴 × 𝐵𝐷)/(𝐴𝐵 × 𝐴𝐷)
𝐵𝐶2/𝐴𝐶2=𝐵𝐷/𝐴𝐷
Hence proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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