Question

AB=CD and angle ABC=BCD prove that I)AC=BD and ii)BE=CE​

Answer 1

$\huge{\boxed{\red{\bf{Question}}}}$

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that

(i) ΔAMC ≌ ΔBMD

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Ans. ∵ M is the mid-point of AB.

∴ BM = AM

[Given]

$\huge\underline\mathfrak\orange{Step(i)}$

In ΔAMC and ΔBMD, we have

CM = DM

[Given]

AM = BM

[Proved]

∠AMC = ∠BMD

[Vertically opposite angles]

∴ ΔAMC ≌ ΔBMD

[SAS criteria]

$\huge\underline\mathfrak\green{Step(ii)}$

∴ΔAMC ≌ ΔBMD

∴ Their corresponding parts are equal.

⇒ ∠MAC = ∠MBD

But they form a pair of alternate interior angles.

∴ AC || DB

Now, BC is a transversal which intersecting parallel lines AC and DB,

∴ ∠BCA + ∠DBC = 180°

But ∠BCA = 90°

[∴ ΔABC is right angled at C]

∴ 90° + ∠DBC = 180°

or ∠DBC = 180° – 90° = 90°

Thus, ∠DBC = 90°

$\huge\underline\mathfrak\purple{Step(iii)}$

Again, ΔAMC ≌ ΔBMD

[Proved]

∴ AC = BD

[c.p.c.t]

Now, in ΔDBC and ΔACB, we have

∠DBC = ∠ACB

[Each = 90°]

BD = CA

[Proved]

BC = CB

[Common]

∴ Using SAS criteria, we have

ΔDBC ≌ ΔACB

${\huge{\fcolorbox{purple}{WHITE}{♡HAN JU HUI♡}}}$