Give me the best way for solving the question of class 9th 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
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)