Question

In right triangle ABC, right angle is at C, M is the mid-point of hypotenuse AB. Cis
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. triangle AMC congurent to triangle BMD
ii.<DBC is right angle
iii.triangle DBC congurent to triangle ACB
iv.CM=1/2 AB
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Answer 1

Answer:

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Answer 2

Figure:-

Given:-

• In right triangle ABC, right angle is 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.

To Find:-

• i. triangle AMC congurent to triangle BMD
• ii.<DBC is right angle
• iii.triangle DBC congurent to triangle ACB
• iv.CM=1/2 AB

Solutions:-

(i). In ∆AMC and ∆BMD

AM = BM ⠀⠀⠀⠀⠀⠀ (M is the mid point of AB)

<AMC = <BMD⠀⠀⠀(vertical opposite angle)

CM = DM ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(Given)

∆AMC ~ ∆BMD ⠀⠀⠀⠀(By SAS Congruence)

AC = BD ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(By CPCT)

(ii). <ADM = <BDM

<ACM and <BDM are alternate interior angle.

Since, alternate angles are equal,

DB // AC

<DBC + <ACB = 180°⠀⠀ (Co - interior angle)

<DBC + 90° = 180°

<DBC = 180° - 90°

<DBC = 90°

(iii). In ∆DBC and ∆ACB,

DB = AC ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(Already proved)

<DBC = <ACB ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(Each 90°)

BC = CB ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(Common)

∆DBC ~ ∆ACD⠀⠀⠀⠀⠀⠀⠀(SAS Congruence)

(iv). ∆DBC ~ ∆ACD

AB = DC ⠀⠀⠀⠀⠀⠀⠀⠀⠀ (By CPCT)

AB = 2CM

CM = 1/2 AB

Additional Information:-

• Straight line = Angle which measures 180°.
• Supplementary Angle = The two angle are supplementary, if there sum are 90°.
• Complementary Angle = The two angle are complementary, if these sum are 90°.

Some Important:-

A transversal intersects two parallel lines.

• The corresponding angles are equal.
• The vertically opposite angles are equal.
• The alternate interior angles are equal.
• The alternate exterior angles are equal.
• The pair of interior angles on the same side of the transversal is supplementary.