7. In Fig. 6.38, altitudes AD and CE of A ABC
intersect each other at the point P. Show
that:
(i) AAEP - ACDP
(ii) AABD-ACBE
(iii) AAEP-AADB
(iv) A PDC - ABEC
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7. In Fig. 6.38, altitudes AD and CE of A ABC
intersect each other at the point P. Show
that:
(i) AAEP - ACDP
(ii) AABD-ACBE
(iii) AAEP-AADB
(iv) A PDC - ABEC
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A ABC
and, altitude AD and CE of triangle intersects each other at the point P.
To Prove : AAEP - ACDP
Proof:
In AAEP and ΔCD
ZAEP = LCDP ZIP = CODE
Hence, AAEP - ACDP Hence provedA ABC
(Since AD & CE are altitudes) (Vertically opposite angles)
(AA Similarity)In A ABD & ACNE
LABD = ZCBE
JAB = JAB
By using AA similarity
AABD - ABCE
(common angle)
(Since AD & CE are altitudes, both 90°)In AES and AADB
ZPA = DAB
ZAEP = LAB By using AA similarity So, ΔΑΕΡ» ADB
Ch6 10th_6.3.7 (iii).mp4
In AES and AADB
ZPA = DAB
ZAEP = ZADB
(common angle)
(Since AD & CE are altitudes, both 90°)
(common angle)
(Since AD & CE are altitudes, both 90°)ABC and ADC
ZDC = ZPCD
ZEB = ZCDP
By using AA similarity
So, ABEC - APDC
Hence proved
(common angle)
(Since AD & CE are altitudes, both 90°)