Give me the best way for solving the question of class 9^{th} of Areas of Parallelograms and Triangles of exercise 9.4 of math. Give me the easiest way for solving this question in simple way of question no.8(4) In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that: Ncert solutions class 9 chapter 9-38 (iv) ΔFCB ≅ ΔACE

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# In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ^ DE meets BC at Y. Show that: Ncert solutions class 9 chapter 9-38 (iv) ΔFCB ≅ ΔACE Q.8(4)

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We know that each angle of a square is 90°.

∴∠FCA = ∠BCE = 90º

∴∠FCA+∠ACB = ∠BCE+∠ACB

∴∠FCB = ∠ACE

In ∆FCB and ∆ACE,

∠FCB = ∠ACE

FC = AC (Sides of square ACFG)

CB = CE (Sides of square BCED)

∆FCB ≅ ∆ACE (SAS congruency)