triangles

Q) In the given figure, Δ FEC ≅ Δ GDB and ∠ 1 = ∠ 2. Prove that Δ ADE ~ Δ ABC. Ans: Step 1: Since Δ FEC ≅ Δ GDB ∴ BD = CE     (by CPCT) Step 2: Since ∠ 1 = ∠ 2 ∴ AD = AE      (since sides […]

Q) In the given figure, ∆ AHK ~ ∆ ABC. If AK = 8 cm, BC = 3.2 cm and HK = 6.4 cm, then find the length of AC. Ans: Step 1: Since Δ AHK ~ Δ ABC                (given) Therefore, ∴ AC = Step 2:  It is given

Q) Sides AB, BC and the median AD of ∆ ABC are respectively proportional to sides PQ, QR and the median PM of another ∆ PQR. Prove that ∆ ABC ~ ∆ PQR Ans:  Given that, In Δ ABC and Δ PQR, Since AD is median of BC, hence BC = 2BD Similarly, PM is median

Q) E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ ABE ~ ∆ CFB. Ans: Given: ABCD is a parallelogram and line AD is extended to point E. Line BE intersects CD at point F To Prove: △ ABE ~ △ CFB

Q) In 𝛥ABC, D, E and F are midpoints of BC,CA and AB respectively. Prove that △ 𝐹𝐵𝐷 ∼ △ DEF and △ DEF ∼ △ ABC Ans: (i): Prove that △ 𝐹𝐵𝐷 ∼ △ DEF: Let’s start from comparing triangles △ FBD and △ DEF. Since by midpoint theorem, The line segment in a

Q) In 𝛥ABC, P and Q are points on AB and AC respectively such that PQ is parallel to BC. Prove that the median AD drawn from A on BC bisects PQ. Ans: Step 1: Let’s start from comparing triangles △ APR and △ ABD. Here we have: ∠ APR  = ∠ ABD      (corresponding

Q)  In the given figure, EA/EC = EB/ ED , prove that ΔEAB ~ ΔECD Ans: Step 1: It is given that: ∴  by cross multiplying, we can say that: Step 2: Let’s look at Δ EAB and Δ ECD: ∠ AEB = ∠ DEC    (∵ Vertically Opposite Angles) Step 3: Let’s compare Δ EAB