Question

Construct a triangle ABC with BC =6.3 cm. Draw aw the right bisector of side BC of the triangle ABC

Answer 1

Answer:

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Step-by-step explanation:

Step 1: Draw a line segment BC = 3.6 cm. Step 2: At the point B, draw ∠XBC = 60°. Step 3: Draw an arc which intersects XB at point D form point B and with radius 4.8 cm Step 4: Join DC. Step 5: Draw a perpendicular bisector of DC which intersects DB at A. Step 6: Join AC. Hence, △ABC is the required triangle. Read more on Sarthaks.com - https://www.sarthaks.com/616865/construct-a-abc-in-which-bc-3-6-cm-ab-ac-4-8-cm-and-b-60

Answer 2

Given: In triangle ABC, BC=6.3cm, AB=4.2cm and AC=5cm

Construction:

1. Draw line BC = 6.3cm
2. Taking B as center and radius 4.2cm, draw an arc
3. Taking C as center and radius 5cm, draw another arc which intersects the first arc at A.
4. Join AB and AC.
5. Then, △ABC is formed.
6. Again with center B and C and radius greater than  $1/2$ BC. Now, draw arcs which intersects each other at L and M
7. Join LM intersecting AC at D and BC at E
8. Join DB.

Proof:

In △DBE and △DCE

BE=EC (LM is the bisector of BC)

∠DEB=∠DEC (Each $90^0$)

DE=DE (Common)

△DBE≅△DCE [SAS criterion of congruence]

DB=DC [Corresponding Parts of Congruent Triangle (CPCT)]

Therefore, D is equidistant from B and C

The complete question is:

Construct a triangle ABC, in which BC=6.3cm, AB=4.2cm and AC=5cm. Draws perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.

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