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Apollonius's Theorem

In fig., DE B...

Question

In fig., DE || BC. If AD = x, DB = x - 2, AE = x + 2 and EC = x - 1, find the value of x.

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Solution

In $△$ ABC, we have

DE || BC

x = 4

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D and E are points on the sides AB and AC respectively of a $Δ A B C$ such that DE || BC
Find the value of x, when

(i) AD = x cm, DB = (x - 2) cm,

AE = (x + 2) cm and EC = (x - 1) cm.

(ii) AD = 4 cm, DB = (x - 4) cm, AE = 8 cm and EC = (3x - 19) cm.

(iii) AD = (7x - 4) cm, AE = (5x - 2) cm, DB = (3x + 4 ) cm and EC = 3x cm.

D E ∥ B C .

A D = x, D B = x - 2, A E = x + 2

E C = x - 1

x

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