Question

E and F are the mid points of the sides AB and AC, respectively of triangle ABC, CE and BF are produced to X and Y respectively, so that EX=CE and FY=BF, AX and AY are joined. Prove that
1) Triangle AEX is equal and similar to Triangle BEC and Triangle AFY is equal and similar to Triangle CFB
2) XAY is a straight line ​

Answer 1

Answer:

Step-by-step explanation:

In triangles AEX & BEC, we have AE = BE

(since E is the mid-point of AB)

angle AEX = angle BEC (V.O.A) EX = EC (given)  

therefore,  

angle AEX congruent to angle BEX (SAS) => angle XAE = angle CBE ( by CPCT)

 => angle XAB = angle CBA

(since angle XAE = angle XAB & angle CBE = angle CBA)

But, angle XAB & angle CBA are alternate interior angles formed when transversal

AB meets XA at A and BC at B. => XA II BC  ... (1)

Similarly it can be proved that,

 triangle AFY congruent to triangle

CFB & AY II BC  ...(2)

From (1) & (2),we get BC II XA & BC II AY So, XA II AY Hence

, XAY is a straight line