Mathematics

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that :
(i) triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.

Mid-point Theorem

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Answer

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that : Mid-point Theorem, Concise Mathematics Solutions ICSE Class 9.

(i) In △ HEB and △ FHC,

⇒ BE = CF (Since, opposite sides of parallelogram are equal i.e. AB = CD and E and F are mid-points of AB and CD respectively)

⇒ ∠HBE = ∠HFC (Alternate angles are equal)

⇒ ∠EHB = ∠FHC (Vertically opposite angles are equal)

∴ △ HEB ≅ △ FHC (By A.A.S. axiom)

Hence, proved that triangles HEB and FHC are congruent.

(ii) Since,

△ HEB ≅ △ FHC

We know that,

Corresponding parts of congruent triangle are equal.

⇒ EH = CH and BH = FH.

⇒ H is the mid-point of BF and CE.

In △ AGE and △ DGF,

⇒ AE = DF (Since, opposite sides of parallelogram are equal i.e. AB = CD and E and F are mid-points of AB and CD respectively)

⇒ ∠GEA = ∠GDF (Alternate angles are equal)

⇒ ∠AGE = ∠DGF (Vertically opposite angles are equal)

∴ △ AGE ≅ △ DGF (By A.A.S. axiom)

∴ AG = GF and EG = DG [By C.P.C.T.C.]

⇒ G is the mid-point of DE and AF.

In △ ECD,

F and H are mid-points of sides CD and EC respectively.

∴ FH || DE [By mid-point theorem]

⇒ FH || GE

F and G are mid-points of sides CD and ED respectively.

∴ GF || EC [By mid-point theorem]

⇒ GF || EH

Since, opposite sides of quadrilateral GEFH are parallel.

Hence, proved that GEHF is a parallelogram.

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Mathematics

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that :
(i) triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.

Mid-point Theorem

Answer

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BM = MN = NC.

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Mathematics

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that :(i) triangles HEB and FHC are congruent;(ii) GEHF is a parallelogram.

Mid-point Theorem

Answer

Related Questions

In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that :(i) M, A and N are collinear.(ii) A is the mid-point of MN.

In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.

In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet side BC at points M and N respectively. Prove that :BM = MN = NC.

In triangle ABC; M is the mid-point of AB, N is mid-point of AC and D is any point in base BC. Use intercept theorem to show that MN bisects AD.

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively. Prove that :
(i) triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.

In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that :
(i) M, A and N are collinear.
(ii) A is the mid-point of MN.

In triangle ABC, D and E are points on side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC which meet side AC at points F and G respectively. Through F and G, lines are drawn parallel to AB which meet side BC at points M and N respectively. Prove that :
BM = MN = NC.