Triangle ABC and triangle DBC have common base BC . prove that A(∆AB

1.	Triangle ABC and triangle DBC have common base BC . prove that A(∆ABC) upon A(∆DBC)=AO upon DO
Answer» ong>Step-by-step explanation: *it is proved by the property of:* *it is proved by the property of:the ratio of the area of two triangles with equal BASES is equal to the ratio of their CORRESPONDING HEIGHTS.* *i.e. h1/h2 areas* *proportionate* to *heights. AO and* DO *are* *the* *corresponding* *heights* of the *given* *triangles. where* BC is *their* *common* *base.* so it is *proved* *that, A(∆ABC) UPON A(∆DBC)=AO upon DO*

Triangle ABC and triangle DBC have common base BC . prove that A(∆ABC) upon A(∆DBC)=AO upon DO

Answer»

ong>Step-by-step explanation:

it is proved by the property of:

it is proved by the property of:the ratio of the area of two triangles with equal BASES is equal to the ratio of their CORRESPONDING HEIGHTS.

i.e. h1/h2

areas proportionate to heights.

AO and DO are the corresponding heights of the given triangles.

where BC is their common base.

so it is proved that,

A(∆ABC) UPON A(∆DBC)=AO upon DO

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Triangle ABC and triangle DBC have common base BC . prove that A(∆ABC) upon A(∆DBC)=AO upon DO​

i.e. h1/h2

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Triangle ABC and triangle DBC have common base BC . prove that A(∆ABC) upon A(∆DBC)=AO upon DO