You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Figures on the Same Base and Between the Same Parallels

AB is a line ...

Question

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE

Open in App

Solution

It is given that ∠EPA = ∠DPB

⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In DAP and EBP,

∠DAP = ∠EBP (Given)

AP = BP (P is mid-point of AB)

∠DPA = ∠EPB (From above)

∴ ΔDAP ≅ ΔEBP (ASA congruence rule)

∴ AD = BE (By CPCT)

Suggest Corrections

Similar questions

Q. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that

∠ B A D = ∠ A B E

and

∠ E P A = ∠ D P B

. Show that

(i) Δ DAP

≅

Δ EBP

(ii) AD = BE

A B

is a line segment and

P

is its mid-point.

D

and

E

are points on the same side of

A B

such that

∠ B A D = ∠ A B E

and

∠ E P A = ∠ D P B

. Show that

Δ D A P ≅ Δ E B P

$A B$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of
$A B$ such that $∠ B A D = ∠ A B E and ∠ E P A = ∠ D P B$ (See the given figure). Show that
$Δ D A P ≅ Δ E B P$

A B

is a line segment and

P

is its mid-point.

D

and

E

are points on the same side of

A B

such that

∠ B A D = ∠ A B E

and

∠ E P A = ∠ D P B

. Show that
(i)

△ D A P ≅ △ E B P

(ii)

A D = D E

463829_10bf09fbcc1745b8bf4de36486798949.463829-Q

Question 7 (ii)
AB is a line segment and P is its mid-point. D and E are points on the same side of
AB such that $∠ B A D = ∠ A B E and ∠ E P A = ∠ D P B$ (See the given figure). Show that
AD = BE

Join BYJU'S Learning Program

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that (i) ΔDAP ≅ ΔEBP (ii) AD = BE

It is given that ∠EPA = ∠DPB ⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE ⇒ ∠DPA = ∠EPB In DAP and EBP, ∠DAP = ∠EBP (Given) AP = BP (P is mid-point of AB) ∠DPA = ∠EPB (From above) ∴ ΔDAP ≅ ΔEBP (ASA congruence rule) ∴ AD = BE (By CPCT)

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE

It is given that ∠EPA = ∠DPB

⇒ ∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In DAP and EBP,

∠DAP = ∠EBP (Given)

AP = BP (P is mid-point of AB)

∠DPA = ∠EPB (From above)

∴ ΔDAP ≅ ΔEBP (ASA congruence rule)

∴ AD = BE (By CPCT)