Question

The figure shows a rhombus ABCD. The diagonal
DB is produced to E such that BC = BE and
CDE = 46°.
Find
1) BAD
2) BCE
​

Answer 1

Answer:

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Answer 2

Given :-

• ABCD is a rhombus.
• The diagonal
• DB is produced to E such that BC = BE .
• ∠CDE = 46°.

To Find :-

• ∠BAD .
• ∠BCE .

Solution :-

→DC is parallel to AB. (Opposite sides of parallelogram are parallel.)

→ DB is a transversal line.

So,

→ ∠CDB = ∠ABD (Alternate interior angle).

Therefore,

→ ∠ABD = 46° .

Now, in ∆ABD,

→ AB = AD (All sides of Rhombus are Equal in Length.)

So,

→ ∠ADB = ∠DBA . (Angle opposite to Equal sides are Equal).

and, we have ,

→ ∠ABD = 46°

Than,

→ ∠ADB = ∠DBA = 46°

Therefore,

→ ∠DAB + ∠ADB + ∠DBA = 180° (Angle sum property.)

→ ∠DAB + 46° + 46° = 180°

→ ∠DAB + 92° = 180°

→ ∠DAB = 180° - 92°

→ ∠DAB = 88° (Ans.)

______________________

Now,

in ∆DCB, we have ,

→ DC = CB (All sides of Rhombus are Equal in Length.)

So,

→ ∠CDB = ∠DBC . (Angle opposite to Equal sides are Equal).

and, we have ,

→ ∠CDB = 46°

Than,

→ ∠DBC = 46°.

Now, DE is a straight Line. (180°).

Therefore,

→ ∠DBC + ∠CBE = 180° .

→ 46° + ∠CBE = 180°

→ ∠CBE = 180° - 46°

→ ∠CBE = 134°.

Now, in ∆BCE , we have ,

→ BC = BE (given.)

Than,

→ ∠BCE = ∠BEC .

Hence,

→ ∠BCE + ∠BEC + ∠CBE = 180° . (Angle sum property.)

→ ∠BCE + ∠BCE + 134° = 180°

→ 2∠BCE = 180° - 134°

→ 2∠BCE = 46°

→ ∠BCE = 23° . (Ans.)