Question

in the figure abce is a rectangle and abdf is a parallelogram. you are asked to give a geometrical proof of fe = dc using the congruency of triangles. the proof was started in the following way: proof: in △afe and △bdc: (i) ae = bc (opposite sides of a rectangle are equal). (ii) bd = af (opposite sides of a parallelogram are equal). (iii)∠aef = ∠bcd = 90° (right angles of a rectangle). which of the following congruency statements would be an acceptable step (iv) so that you can then say fe = dc (corresponding sides of congruent triangles)? select an answer: (iv) ∴△aef≅△bdc (saa) (iv) ∴△aef≅△bdc (rhs) (iv) ∴△aef≅△bdc (sss) (iv) ∴△aef≅△bdc (asa)

Explanation:

Step1: Analyze given information

We have two - right - angled triangles $\triangle AEF$ and $\triangle BDC$. We know two pairs of corresponding sides are equal ($AE = BC$ and $BD=AF$) and one pair of corresponding right - angles are equal ($\angle AEF=\angle BCD = 90^{\circ}$).

Step2: Recall congruence criteria

The RHS (Right - Hypotenuse - Side) congruence criterion states that if the hypotenuse and one side of a right - angled triangle are equal to the hypotenuse and one side of another right - angled triangle, then the two triangles are congruent. In $\triangle AEF$ and $\triangle BDC$, we have right - angles at $\angle AEF$ and $\angle BCD$, and two pairs of corresponding sides.

Answer:

(iv) $\therefore\triangle AEF\cong\triangle BDC$ (RHS)