In the adjoining figure, ABCD is a parallelogram. E and F are mid-points of the sides AB and CD respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that
(i) △HEB ≅ △HCF
(ii) GEHF is a parallelogram.

Question

In the adjoining figure, ABCD is a parallelogram. E and F are mid-points of the sides AB and CD respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that
(i) △HEB ≅ △HCF
(ii) GEHF is a parallelogram.

KnowledgeBoat · Accepted Answer

(i) We know that, ABCD is a parallelogram, ∴ FC || BE ∠CEB = ∠FCE (Alternate angles) ⇒ ∠HEB = ∠FCH .......(1) ∠EBF = ∠CFB (Alternate angles) ⇒ ∠EBH = ∠CFH .......(2) Here E and F are mid-points of AB and CD BE = AB ........(3) CF = CD ........(4) We know that ABCD is a parallelogram, AB = CD Now dividing both sides by AB = CD Using equations 3 and 4 we get, BE = CF .......(5) In △HEB and △HCF, ∠HEB = ∠FCH (Using eqn. i) ∠EBH = ∠CFH (Using eqn. ii) BE = CF (Using eqn. v) ∴ △HEB ≅ △HCF (By ASA axiom of congruency) Hence, proved that △HEB ≅ △HCF. (ii) AB = CD (As ABCD is a parallelogram) Hence, AE = CF (As E and F are mid-points of the sides AB and CD respectively) As, AB || CF we can say that, AE || CF Since, AE = CF and AE || CF ∴ AECF is a parallelogram. ∴ AF || EC From figure we get, GF || EH ........(1) AB = CD (As ABCD is a parallelogram) Hence, DF = EB (As E and F are mid-points of the sides AB and CD respectively) and DF || EB. Since, DF = EB and DF || EB ∴ DEBF is a parallelogram. ∴ DE || FB From figure we get, GE || FH ........(2) From 1 and 2 we get, GF || EH and GE || FH. ∴ GEHF is a parallelogram. Hence, proved that GEHF is a parallelogram.

Mathematics

In the adjoining figure, ABCD is a parallelogram. E and F are mid-points of the sides AB and CD respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that
(i) △HEB ≅ △HCF
(ii) GEHF is a parallelogram.

Mid-point Theorem

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Mathematics

In the adjoining figure, ABCD is a parallelogram. E and F are mid-points of the sides AB and CD respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that(i) △HEB ≅ △HCF(ii) GEHF is a parallelogram.

Mid-point Theorem

Related Questions

Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.

ABC is an isosceles triangle with AB = AC. D, E and F are mid-points of the sides BC, AB and AC respectively. Prove that line segment AD is perpendicular to EF and is bisected by it.

In the adjoining figure, ABCD is a parallelogram. E and F are mid-points of the sides AB and CD respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that
(i) △HEB ≅ △HCF
(ii) GEHF is a parallelogram.