Connect KC, Quadrilateral AEDC is cyclic => ∠ (BED)= ∠ (ACB) .. ( both angles supplement to ∠ (AED)) we have ∠ (BAC)= ∠ (BKC) => ∠ (ABF)= ∠ (KBC) so triangle BEG similar to BCM and triangles BEH similar to BCK …( case AA) and we have BG/BM=BE/BC=BH/BK so BG/BH=BM/BK and GM//HK
Connect KC,
ReplyDeleteQuadrilateral AEDC is cyclic => ∠ (BED)= ∠ (ACB) .. ( both angles supplement to ∠ (AED))
we have ∠ (BAC)= ∠ (BKC) => ∠ (ABF)= ∠ (KBC)
so triangle BEG similar to BCM and triangles BEH similar to BCK …( case AA)
and we have BG/BM=BE/BC=BH/BK
so BG/BH=BM/BK
and GM//HK
add sketch of problem 1248
ReplyDeletehttps://photos.app.goo.gl/2rCpcFijxxBSHph78