E and F are midpoints of sides AB,AC of triangle ABC,CE and BF are produced to X,Y respectively,so that EX=CE and FY=BF,AX and AY are joined.Find a triangle congruent to triangle AEX and demonstrate the congruency.Prove that XAY is a straight line.
Answers
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.
Step-by-step explanation:
in triangle ACX
EF || AX
EF=1/2AX
in triangle ABY
EF||AY
EF=1/2AY
so, AX||AY
this is possible only when XAY in a straight line
Hence proved